Codeforces 1165F2（二分内的check）

要点

• 二分答案，内部喜闻乐见的拖延策略：对于某个打折玩具，就选最晚的打折时间买，答案并不会变劣，只是购买时间的平移。
• 注意最晚时间不是预处理的东西，是二分内部的、在mid以内的最晚时间。

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 2e5 + 5;
int n, m, k[maxn], d[maxn], t[maxn];
int Late[maxn], sum;

bool ok(int mid) {
    for (int i = 1; i <= n; i++) {
        Late[i] = 0;
    }
    vector<int> day[maxn];
    for (int i = 1; i <= m; i++) {
        if (d[i] <= mid)
            Late[t[i]] = max(Late[t[i]], d[i]);
    }
    for (int i = 1; i <= n; i++) {
        if (k[i] && Late[i])
            day[Late[i]].push_back(i);
    }

    int s = sum, cur = 0;
    for (int i = 1; i <= mid; i++) {
        cur++;
        if (i < maxn && day[i].size()) {
            for (int t : day[i]) {
                if (cur >= k[t]) {
                    cur -= k[t];
                    s -= k[t];
                } else {
                    s -= cur;
                    cur = 0;
                    break;
                }
            }
        }
    }
    return cur >= s * 2;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &k[i]), sum += k[i];
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &d[i], &t[i]);
    }

    int l = 1, r = 4e5 + 5, ans;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (ok(mid))    ans = mid, r = mid - 1;
        else    l = mid + 1;
    }
    return !printf("%d\n", ans);
}